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copied!<p>The code is right. <code>auto&& p = expr</code> means the type of <code>p</code> is <code>T&&</code> where <code>T</code> will be inferred from <code>expr</code>. The <code>&&</code> here indicates a rvalue reference, so e.g.</p> <pre><code>auto&& p = 1; </code></pre> <p>will infer <code>T == int</code> and thus the type of <code>p</code> is <code>int&&</code>.</p> <p>However, references can be collapsed according to the rule:</p> <pre><code>T& & == T& T& && == T& T&& & == T& T&& && == T&& </code></pre> <p>(This feature is used to implement perfect forwarding in C++11.)</p> <p>In the case</p> <pre><code>auto&& p = x; </code></pre> <p>as <code>x</code> is an lvalue, an rvalue reference cannot be bound to it, but if we infer <code>T = int&</code> then the type of <code>p</code> will become <code>int& && = int&</code>, which is an lvalue reference, which can be bound to <code>x</code>. Only in this case <code>auto&&</code> and <code>auto&</code> give the same result. These two are different though, e.g.</p> <pre><code>auto& p = std::move(x); </code></pre> <p>is incorrect because <code>std::move(x)</code> is an rvalue, and the lvalue reference cannot be bound to it.</p> <p>Please read <a href="http://thbecker.net/articles/rvalue_references/section_01.html">C++ Rvalue References Explained </a> for a walk through.</p>

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